2y+5+2y+5+y^2-62=360

Solution for 2y+5+2y+5+y^2-62=360 equation:

2y+5+2y+5+y^2-62=360

We move all terms to the left:

2y+5+2y+5+y^2-62-(360)=0

We add all the numbers together, and all the variables

y^2+4y-412=0

a = 1; b = 4; c = -412;
Δ = b2-4ac
Δ = 42-4·1·(-412)
Δ = 1664
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{1664}=\sqrt{64*26}=\sqrt{64}*\sqrt{26}=8\sqrt{26}$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-8\sqrt{26}}{2*1}=\frac{-4-8\sqrt{26}}{2}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+8\sqrt{26}}{2*1}=\frac{-4+8\sqrt{26}}{2}
$